Question

Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

Answer 1

This calls for the relationship of distance equals speed or rate times the time.
d=rt
You are given their respective speeds, and the different start times. They will catch up when each has traveled the same distance. So distance will be equal. Set up and equation for each getting their distance and then equate the two equations.

Answer 2

#1 speed is traveling at 6 mph
let his time be t. His distance will be 6t. Clear so far??

Answer 3

yes

Answer 4

#2 speed is 10 mph
his time is 3 hours less than #1, (He left 3 hrs after #1)
his distance will be 10(t-3)

Answer 5

When the distances are equal they have met.
6t=10(t-3)
6t=10t-30
-4t=-30
t=30/4=7.5 hrs
#1 travels 7.5 * 6=45 mi.
#2 travels 4.5*10 =45 mi
It checks.

Answer 6

7.5 hours after the first cyclist left, they meet up.

Answer 7

Did you follow with understanding?

Answer 8

yes,thanku...